# Full text of "Treatise On Analysis Vol-Ii"

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```11    THE SPECTRAL THEORY OF HUBERT        409

eigenvector of £/, then (/®/)~ is an eigenvector of the unitary operator corresponding
to u x w, for the eigenvalue 1. To prove that (y) implies (a), use the last criterion of
(b); if (/|T) = 0, introduce the measure v^m~ftj (15.11.1), and observe that this
measure on U is diffuse (15.11.6). Then we are reduced to proving that

u

1 dv(Q

= 0.

Write this relation in the form

lim JJ (~J

and remark that the diagonal of U x U is (v ® v)-negligible.)

(e)    With the notation of Section 13.9, Problem 13(c), show that if 6 is irrational, the
mapping z\-+e**i9z of U onto U is not weakly mixing. (Calculate the spectrum of the
corresponding unitary operator £/.)

(f)    Suppose that the space L£(X, p.) is the Hilbert sum of the subspace C • 1 (the
classes of the constant functions) and an at most denumerable family (H/X/6j of
Hilbert spaces, where each H, has a  Hilbert basis (enJ\eZ such that U-enJ*=
en + ! , j for all n e Z (cf . Problem 1 5). Then the mapping u is mixing. In particular, if.
(X, JLA, p) is the Bernoulli scheme B(i, J) (Section 13.21, Problem 18) then u is mixing.
(If, for each neZ,fn is the function on lz such that fn(x) « — 1 if prnx = 0, and
/„(*) = 1 if prn x = 1, then the classes of the finite products /ni/«2 • • •/„*, in which all
the indices are distinct, form together with the class of 1 an orthonormal basis of
Lc(X, /A) Likewise show that, if X is the torus T2, TT : R->T the canonical homo-
morphism and /x the normalized Haar measure on X, then u defined by

\ rr(x + 2y))

is mixing.

(g) Suppose that u is ergodic with respect to /x, so that (by virtue of (c)) the eigen-
values of U form an at most denumerable subgroup G of U, the eigenspace correspon-
ding to an eigenvalue a e G being a line D(a) in L£(X, /A). Show that there exists a
family of eigenvectors A e D(a) of U such that |/« | = 1 almost everywhere in X and
such that/a/, = /a/0 almost everywhere, for all pairs (a, ]3) of points of G. (Let ha e D(a)
be such that |/z«| = 1 almost everywhere, for all aeG; for each pair of points
a, |3 e G we may write ha0 = r(oc, /3)hah0 almost everywhere, where r(a, fi) e U is a
constant. Denote by A the subgroup of Ux generated by the ha for a e G and the
group of constant functions from X to U (which may be identified with U). Show that
there exists a homomorphism 6 : A-*U such that 0(0 = £ fc>r all £ eU. For this
purpose, arrange the h* in a sequence (hn) and proceed by induction: if 0 is already
defined on the subgroup A« generated by U and the hj such thaty"</z, distinguish
two cases according as h* is not constant almost everywhere for any nonzero k e Z,
or on the contrary that there exists a smallest positive integer k such that /zf is constant
almost everywhere; use the fact that for all j e U and all integers k > 0, there exists
T] e U such that yk = £. Then take/« = 0(/za)/za.)

(h) Let v : X -> X be another mapping which is ergodic with respect to /x, and let
V be the corresponding unitary operator. Show that if u satisfies the hypotheses of (g)
and if U, V have the same eigenvalues, then u and t; are conjugate (Section 13.12,
Problem 11).= 1 for all AeK. Then there exists no operator T'e &(E)
```